# Q. The length of a diagonal of a regular pentagon of side length 1 is `x` units. Then sin 54° = ? A) `(2x)/5` B) `x/2` C) `(sqrt3 + 1)/(2sqrt3) x` D) `(1/sqrt3)x`

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Draw regular pentagon ABCDE with AB=BC=CD=DE=EA=1. Draw AC with length x. (AC is an arbitrary diagonal -- any diagonal will do.)

Since the pentagon is regular, every exterior angle is `360/5=72^@` which implies that every interior angle has measure `108^@` ; in particular `m/_B=108^@` .

Draw the altitude from B to AC and label the foot F. Then `m/_FBC=54^@` . (Note that triangle ABC is isosceles -- the altitude drawn from the vertex angle of an isosceles triangle is also the median, the perpendicular bisector of the base, and the angle bisector of the vertex angle.)

Triangle FBC is a right triangle. By the definition of the sine of an angle, `sinalpha="sideopposite"/"hypotenuse"` we can find the sine of 54 degrees.

The hypotenuse is the length of a side of the pentagon which is 1. The side opposite is FC which has length `x/2` .

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`sin54^@=x/2` or 1/2 the length of the diagonal of a regular pentagon.

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